The number of bacteria N at a certain time t is given by N = 100et / 10, when there are 100 bacteria at time t = 0 hours. Find how many hours t it takes for the bacteria to double (that is, increase from 100 to 200 bacteria). [Use either ln (2) ≈ 0.7 to approximate your answer as appropriate.]

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Mathematics

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1 March, 05:43

The number of bacteria N at a certain time t is given by N = 100et / 10, when there are 100 bacteria at time t = 0 hours. Find how many hours t it takes for the bacteria to double (that is, increase from 100 to 200 bacteria). [Use either ln (2) ≈ 0.7 to approximate your answer as appropriate.]

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Tarzan · Answer 1

I assume you meant N=100e^ (t/10) so

200=100e^ (t/10)

2=e^ (t/10) taking the natural log of both sides

ln2=t/10

t=10ln2

t≈6.93 hours

check ...

N=100e^ (.693) ≈200